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Finding Basis of Null Space and Range

  1. Feb 15, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove T is a linear transformation and find bases for both N(T) and R(T).


    2. Relevant equations



    3. The attempt at a solution


    T:M2x3(F) [itex]\rightarrow[/itex] M2x2(F) defined by:

    T(a11 a12 a13)
    (a21 a22 a23)

    (this is one matrix)
    =

    (2a11-a12 a13+2a12)
    ( 0 0)

    (this is one matrix)

    So I verified that it is a linear transformation by checking that T(cx+y)=cT(x)+T(y). But I don't understand how to find a basis for the null space and range.

    I can see that since N(T)={x:T(x)=0} that N(T) here it all vectors of the form:

    (t/2 t -2t)
    ( b b b)

    (this is one matrix)

    Since the 2nd row in our domain always goes to 0, the second row is arbitrary, which I represented by b.

    How do I find a basis for all multiples of the matrix

    t(1/2 1 -2)
    ( b b b)?

    And I'm not even sure on how to start off finding the basis for the range. All help is appreciated. Thanks!
     
  2. jcsd
  3. Feb 15, 2012 #2

    Deveno

    User Avatar
    Science Advisor

    do you have any idea of what the dimension of M2x3(F) is?

    can you think of a basis for it?

    can you express N(T) in terms of this basis?

    alternately: do you know how to "vectorize" a matrix?

    suppose you "vectorized" the elements of M2x3(F) and M2x2(F). could you pick bases for these spaces, and calculate a matrix for T in terms of those bases? what size would it have? could you find its rank and nullity?
     
  4. Feb 15, 2012 #3
    do you have any idea of what the dimension of M2x3(F) is?

    Since it is 2x3, the dimension would be 6?

    Because we would need 6 vectors:

    \begin{array}{ccc}
    1 & 0 & 0 \\
    0 & 0 & 0 \end{array},

    \begin{array}{ccc}
    0 & 0 & 0 \\
    1 & 0 & 0 \end{array},

    \begin{array}{ccc}
    0 & 1 & 0 \\
    0 & 0 & 0 \end{array}

    \begin{array}{ccc}
    0 & 0 & 0 \\
    0 & 1 & 0 \end{array},

    \begin{array}{ccc}
    0 & 0 & 1 \\
    0 & 0 & 0 \end{array}

    \begin{array}{ccc}
    0 & 0 & 0 \\
    0 & 0 & 1 \end{array}

    N(T) in terms of this basis would be multiples of:

    (1/2)(first matrix)+(third matrix)+(-2)(fifth matrix)?

    Something doesn't seem right to be hmm.

    Also, what do you mean by vectorizing?
     
  5. Feb 15, 2012 #4

    vela

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    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Good start. But what about the second matrix? Doesn't T map that to 0 as well? What about the other basis elements?

    You can express any matrix in M2x3 as a linear combination of those six matrices, so if you label those six matrices e1, … , e6, you could write
    $$\begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{pmatrix} = a_{11}\mathbf{e}_1 + a_{21}\mathbf{e}_2 + a_{12}\mathbf{e}_3 + a_{22}\mathbf{e}_4 + a_{13}\mathbf{e}_5 + a_{23}\mathbf{e}_6$$ This matrix would therefore correspond to the coordinate vector ##(a_{11}, a_{21}, a_{12}, a_{22}, a_{13}, a_{23})^T##.
     
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